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2 years ago

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A 92.0-kg skydiver falls straight downward with an open parachute through a vertical height of 325 m. The skydiver’s velocity re

mains constant. What is the work done by the nonconservative force of air resistance, which is the only nonconservative force acting?

1 answer:

Alborosie2 years ago

5 0

Answer:

The work done is $293.0kJ$

Explanation:

<u>Data</u>

Mass: $m=92.0kg$

height: $h=325m$

acceleration due to gravity: $g=9.81m/s^{2}$

Non conservative force act

The work is:

$W=$ Δ $KE$ +Δ $PE$

KE and PE means kinetic energy and potential energy respectively.

The change in kinetic energy is 0, since the skydiver’s velocity remains constant.

Therefore

$W=0+$ Δ $PE$

$W=$ Δ $PE$

$W=mgh$

substitute the values

$W=92.0kg*9.81m/s^{2} *325m\\W=293020J\\W=293.0kJ$

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A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha

Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

$PQ = \sqrt{4^{2}+3^{2}}=5 m$

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

$E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C$

Electric field at P due to the charge q is

$E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C$

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

$E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C$

3 0

2 years ago

In the context of energy transfers with hot and cold reservoirs, the sign convention is that _______________.

Likurg_2 [28]

Answer:

B. QC > 0; QH < 0

Explanation:

Given that there are two reservoir of energy.

Sign convention for heat and work :

1.If the heat is adding to the system then it is taken as positive and if heat is going out from the system then it is taken as negative.

2. If the work is done on the system then it is taken as negative and if the work is done by the system then it is taken as positive.

From hot reservoir heat is going out that is why it is taken as negative

$Q_H$

From cold reservoir heat is coming inside the reservoir that is why it is taken as positive

$Q_C>0$

That is why the answer will be

$Q_H$ , $Q_C>0$

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2 years ago

A mass on the end of a spring undergoes simple harmonic motion. At the instant when the mass is at its maximum displacement from

hodyreva [135]

Answer:

C. At maximum displacement, its instantaneous acceleration is also at maximum.

Explanation:

Lets take

The general equation of the SHM

Displacement

x= A sinω t

velocity

V= Aω cosω t

Acceleration

a= -Aω² sinω t

Form the above we can say that displacement and the acceleration are in the same phase.

Therefore when displacement is maximum then acceleration also will be maximum.

Therefore the answer is C.

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2 years ago

A rocket is sitting on the launch pad. The engines ignite, and the rocket begins to rise straight upward, picking up speed as it

grandymaker [24]

Answer:

e. Only the rocket's motion from the top of its flight path until just before landing is free-fall.

Explanation:

A free-fall is a fall just under force of gravity. The rocket;s upward motion is result of engine push - even if it was shut down - and rocket free of engine push effect when it reaches it's maximum height after shutting down of engine. Then rockets stops at it's maximum height for a moment and rtuens back as free fall with only force of gravitation pulling it back to ground with acceleration 'g'.

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To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following general equation y(

NeX [460]

Answer:

The transverse displacement is $y(1.51 , 0.150) = 0.055 m$

Explanation:

From the question we are told that

The generally equation for the mechanical wave is

$y(x,t) = Acos (kx -wt)$

The speed of the transverse wave is $v = 8.25 \ m/s$

The amplitude of the transverse wave is $A = 5.50 *10^{-2} m$

The wavelength of the transverse wave is $\lambda = 0540 m$

At t= 0.150s , x = 1.51 m

The angular frequency of the wave is mathematically represented as

$w = vk$

Substituting values

$w = 8.25 * 11.64$

$w = 96.03 \ rad/s$

The propagation constant k is mathematically represented as

$k = \frac{2 \pi}{\lambda}$

Substituting values

$k = \frac{2 * 3.142}{0. 540}$

$k =11.64 m^{-1}$

Substituting values into the equation for mechanical waves

$y(1.51 , 0.150) = (5.50*10^{-2} ) cos ((11.64 * 1.151 ) - (96.03 * 0.150))$

$y(1.51 , 0.150) = 0.055 m$

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2 years ago